Find the partial fraction decomposition of $\frac { 4 } { x ^ { 3 } - 4 x ^ { 2 } + 4 x }$

A) $- \frac { 1 } { 4 x } + \frac { 1 } { 8 ( x - 2 ) } + \frac { 1 } { 8 ( x + 2 ) }$

B) $\frac { 1 } { 2 x } - \frac { 2 } { ( x - 2 ) } + \frac { 4 } { ( x - 2 ) ^ { 2 } }$

C) $- \frac { 1 } { x } + \frac { 2 } { ( x - 2 ) } - \frac { 1 } { ( x + 2 ) }$

D) $\frac { 1 } { x } - \frac { 1 } { ( x - 2 ) } + \frac { 2 } { ( x - 2 ) ^ { 2 } }$

E)none

Question

Find the partial fraction decomposition of $\frac { 4 } { x ^ { 3 } - 4 x ^ { 2 } + 4 x }$

A) $- \frac { 1 } { 4 x } + \frac { 1 } { 8 ( x - 2 ) } + \frac { 1 } { 8 ( x + 2 ) }$

B) $\frac { 1 } { 2 x } - \frac { 2 } { ( x - 2 ) } + \frac { 4 } { ( x - 2 ) ^ { 2 } }$

C) $- \frac { 1 } { x } + \frac { 2 } { ( x - 2 ) } - \frac { 1 } { ( x + 2 ) }$

D) $\frac { 1 } { x } - \frac { 1 } { ( x - 2 ) } + \frac { 2 } { ( x - 2 ) ^ { 2 } }$

E)none

Quizplus · Accepted Answer

First, factor the denominator: 
$$x^3-4x^2+4x=x(x^2-4x+4)=x(x-2)^2$$ 
Next, use partial fraction decomposition to express $\frac{4}{x(x-2)^2}$ as: 
$$\frac{4}{x(x-2)^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$ 
Multiplying both sides of the equation by $x(x-2)^2$ and then simplifying, we get: 
$$4 = A(x-2)^2 + Bx(x-2) + Cx$$ 
To solve for $A$, we can substitute $x=0$ to obtain $4 = A(-2)^2$ and so $A = 1$. 
To solve for $C$, we can substitute $x=2$ to obtain $4 = C(2)$. Thus, $C=2$. 
For $B$, we can substitute $x=1$ to obtain $4 = B(1)(-1)$ and so $B=-4$. 
Therefore, we can rewrite $\frac{4}{x(x-2)^2}$ as: 
$$\frac{4}{x(x-2)^2}=\frac{1}{x}-\frac{4}{x-2}+\frac{2}{(x-2)^2}$$ 
which matches the answer choice D.

Find the Partial Fraction Decomposition Of 4x3−4x2+4x\frac { 4 } { x ^ { 3 } - 4 x ^ { 2 } + 4 x }x3−4x2+4x4​

Find the Partial Fraction Decomposition Of $\frac { 4 } { x ^ { 3 } - 4 x ^ { 2 } + 4 x }$